Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $q = \dfrac{9y^2 - 9y}{y + 2} \times \dfrac{6y + 24}{y^2 + 3y - 4} $
Solution: First factor the quadratic. $q = \dfrac{9y^2 - 9y}{y + 2} \times \dfrac{6y + 24}{(y - 1)(y + 4)} $ Then factor out any other terms. $q = \dfrac{9y(y - 1)}{y + 2} \times \dfrac{6(y + 4)}{(y - 1)(y + 4)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ 9y(y - 1) \times 6(y + 4) } { (y + 2) \times (y - 1)(y + 4) } $ $q = \dfrac{ 54y(y - 1)(y + 4)}{ (y + 2)(y - 1)(y + 4)} $ Notice that $(y + 4)$ and $(y - 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 54y\cancel{(y - 1)}(y + 4)}{ (y + 2)\cancel{(y - 1)}(y + 4)} $ We are dividing by $y - 1$ , so $y - 1 \neq 0$ Therefore, $y \neq 1$ $q = \dfrac{ 54y\cancel{(y - 1)}\cancel{(y + 4)}}{ (y + 2)\cancel{(y - 1)}\cancel{(y + 4)}} $ We are dividing by $y + 4$ , so $y + 4 \neq 0$ Therefore, $y \neq -4$ $q = \dfrac{54y}{y + 2} ; \space y \neq 1 ; \space y \neq -4 $